Solve for $x$ : $ 3|x + 2| + 7 = 6|x + 2| + 5 $
Subtract $ {3|x + 2|} $ from both sides: $ \begin{eqnarray} 3|x + 2| + 7 &=& 6|x + 2| + 5 \\ \\ {- 3|x + 2|} && {- 3|x + 2|} \\ \\ 7 &=& 3|x + 2| + 5 \end{eqnarray} $ Subtract $5$ from both sides: $ \begin{eqnarray} 7 &=& 3|x + 2| + 5 \\ \\ {- 5} && {- 5} \\ \\ 2 &=& 3|x + 2| \end{eqnarray} $ Divide both sides by ${3}$ $ \dfrac{2} {{3}} = \dfrac{3|x + 2|} {{3}} $ Simplify: $ \dfrac{2}{3} = |x + 2| $ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ -\dfrac{2}{3} = x + 2 $ or $ \dfrac{2}{3} = x + 2 $ Solve for the solution where $x + 2$ is negative: $ - \dfrac{2}{3} = x + 2$ Subtract ${2}$ from both sides: $ \begin{eqnarray} - \dfrac{2}{3} &=& x + 2 \\ \\ {- 2} && {- 2} \\ \\ -\dfrac{2}{3} - 2 &=& x \end{eqnarray} $ Change the ${ - 2}$ to an equivalent fraction with a denominator of $3$ $ - \dfrac{2}{3} {- \dfrac{6}{3}} = x $ $ -\dfrac{8}{3} = x $ Then calculate the solution where $x + 2$ is positive: $ \dfrac{2}{3} = x + 2 $ Subtract ${2}$ from both sides: $ \begin{eqnarray} \dfrac{2}{3} &=& x + 2 \\ \\ {- 2} && {- 2} \\ \\ \dfrac{2}{3} - 2 &=& x \end{eqnarray} $ Change the ${ - 2}$ to an equivalent fraction with a denominator of $3$ $ \dfrac{2}{3} {- \dfrac{6}{3}} = x $ $ -\dfrac{4}{3} = x $ Thus, the correct answer is $x = -\dfrac{8}{3} $ or $x = -\dfrac{4}{3} $.